Question

(1) Figure shows a series LCR circuit connected to a variable frequency $230V$ source. $L=5.0H,C=80\mu F,R=40\Omega$ Determine the source frequency which drives the circuit in resonance.

(2) Figure shows a series LCR circuit connected to a variable frequency $230V$ source. $L=5.0H,C=80\mu F,R=40\Omega$ Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.


(3) Figure shows a series LCR circuit connected to a variable frequency $230V$ source. $L=5.0H,C=80\mu F,R=40\Omega$ Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency.

Answer 1

(1) Given,
$L=5.0H$

$C=80\mu F=80\times {10}^{-6}F$
$R=40\Omega$

We know in resonating condition of circuit,

Frequency of source = Natural frequency of AC circuit

In resonance condition
$X_L=X_C$

${\omega }_{r}L=\frac{1}{{\omega }_{r}C}$

$\Rightarrow {\omega }_r=\frac{1}{\sqrt{LC}}$

Putting values, we get
${\omega }_r=\frac{1}{\sqrt{5\times 80\times {10}^{-6}}}$

${\omega }_r=50rad/s$

Final Answer: $50rad/s$

(2) Step 1: Find impedance at resonant frequency
Given,
$L=5.0H,R=40\Omega$
$C=80\mu F=80\times {10}^{-6}F$
$V_{rms}=230V$

Impedance of LCR circuit

$Z=\sqrt{(X_L-X_C{)}^2+R^2}$

In resonance condition
$\Rightarrow X_L=X_C$

Then the impedance of the circuit at resonance,
$\Rightarrow Z=R=40\Omega$ (Purely resistive)

In purely resistive circuit, $\varphi =0^{\circ }$

Step 2: Find current at resonant frequency

Current in the circuit: $I_{rms}=\frac{V_{rms}}{R}$

$I_{rms}=\frac{230}{40}=5.75A$
At resonance,
$I_m=\sqrt{2}{I}_{rms}$

$I_m=5.75\sqrt{2}=8.13A$
Final Answer: $40\Omega ,8.13A$

(3) Step 1: Find RMS current

Given, $L=5.0H,R=40\Omega$
$C=80\mu F=80\times {10}^{-6}F$
$V_{rms}=230V$
At resonant frequency effect of $L$ and $C$ nullify each other. Hence
$I_{rms}=\frac{V_{rms}}{Z}=5.75A$

Step 2: Find RMS potential drop across each element

Potential drop across Inductance:
$V_L=I_{rms}{X}_L=I_{rms}\omega L$

$=5.75\times 50\times 5=1437.5V$

Potential drop across capacitor:
$V_C=I_{rms}{X}_C=\frac{I_{rms}}{\omega C}$

$=\frac{5.75}{50\times 80\times {10}^{-6}}=1437.5V$

Where, $V_L$ and $V_C$ are on opposite phase.

Potential drop across resistor:
$V_R=I_{rms}R=5.75\times 40$

$=230V$

Step 3: Find voltage drop across LC combination

From phasor diagram, potential drop across LC:

$V_{LC}=V_L-V_C$

$V_{LC}=I_{rms}(X_L-X_C)$

At resonance $X_l=X_c$

$\Rightarrow V_{LC}=0$

Final Answer: $V_L=1437.5V,V_C=1437.5V{V}_R=230V,V_{LC}=0$